Continuity

Much of limit analysis relates to a concept known as continuity.
A function is said to be continuous on an interval when the
function is defined at every point on that interval and undergoes no
interruptions, jumps, or breaks. If some function f(x) satisfies these criteria
from x=a to x=b, for example, we say that f(x) is continuous on the interval
[a, b]. The brackets mean that the interval is closed -- that
it includes the endpoints a and b. In other words, that the interval is defined as
a ≤ x ≤ b. An open interval (a, b), on the other hand, would
not include endpoints a and b, and would be defined as a < x < b.

For an example of continuity, start a new worksheet called 02-Continuity, then
recreate the following graph using the provided code. (The f(x)=... is superimposed).

1) Plot x2 from x=0 to x=1
2) Plot -x+2 from x=1 to x=2
3) Plot x2-3*x+2 from x=2 to x=3
4) Create a black point at (0, 0)
5) Create another black point at (3, 2)
6) Combine the plots and points into a single graph with the given bounds

The function f(x) in the graph is known as a piecewise function,
or one that has multiple, well, pieces. As you can see, the function travels
from x=0 to x=3 without interruption, and since the two endpoints are closed
(designated by the filled-in black circles), f(x) is continuous on the closed
interval [0, 3]. To think of it another way, if you can trace a function on an
interval without picking up your pen (and without running over any holes), the
function is continuous on that interval.

Using our knowledge of limits from the previous lesson, we can say that:

and

The Intermediate Value Theorem

One of the more important theorems relating to continuous functions is
the Intermediate Value Theorem, which states that if a function
f is continuous on a closed interval [a, b] and k is any number between f(a) and
f(b), then there must exist at least one number c such that f(c) = k. In other
words, if f is continuous on [a, b], it must pass through every y-value bounded
by f(a) and f(b). In the continuous function graphed above, for example,
f(0) = 0 and f(3) = 2, so f(x) must pass through all y-values bounded by and
including 0 and 2 on the interval [0, 3], which as one can see, it does.

Look at this example, now, of a function that is not continuous on the
interval for which it is shown.

1) Plot 2*x from x=0 to x=1
2) Plot -x+3 from x=1 to x=2
3) Plot -(x-3)3+2 from x=2 to x=3
4) Create a dashed line to indicate that the function jumps to a y-value of 3
when x is equal to 1
5-10) Create open (faceted) or closed (filled-in) points to indicate whether the intervals are open
or closed
11) Combine the plots, line, and points into a graph with the given bounds

The function shown in the graph is not continuous on the closed interval
[0, 3], since it has discontinuities at both x=1 and x=2. A
discontinuity is any x-value at which a function has an interruption, break or
jump -- something that would require you to pick up your pen if you were tracing
the function. The filled-in black circles, again, indicate that
the interval includes that point, while the open circles indicate that the
interval excludes that point. The dashed line at x=1 shows that f(1) = 3, not 2.

Since the above graph has at least one discontinuity, it does not satisfy the
requirements of the Intermediate Value Theorem and therefore does not
have to pass through every y-value between f(0) and f(3).

Even though f(1) = 3, however, the limit of f(x) as x approaches 1 does not
equal 3, since the function approaches a value of 2 from both sides of x=1. But
what about the limit of f(x) as x approaches 2? The function does not approach
one specific value from either side of x=2, but we can still describe its behavior
from the right or the left. This is the basis of one-sided limits, which is the
topic of the next section.

Practice Problems

Determine which of the following functions are continuous on the closed intervals
for which they are shown. For functions that are not continuous, determine the
x-coordinates of their discontinuities.
1)

*As a SageMath-based way of checking your answers, you can use solve(...==..., k)
to solve the given equation for k. For example, to solve for x when the expression
x2-4*x+4 is equal to 0, you would use: